In a double replacement reaction between silver nitrate and magnesium bromide. Calculate the mass of silver bromide which should be produced from 22.5g of silver nitrate. The actual yield was 22.0g, how do you calculate percent yield?
1 Answer
Theoretical yield:
Percent yield:
Explanation:
Start by writing the balanced chemical equation for this double replacement reaction
#2"AgNO"_text(3(aq]) + "MgBr"_text(2(aq]) -> color(red)(2)"AgBr"_text((s]) darr + "Mg"("NO"_3)_text(2(aq])#
Notice that you have a
However, keep in mind that this is true for a reaction that has a
If the percent yield of the reaction is smaller than
So, how many moles of silver nitrate do you have in
#22.5color(red)(cancel(color(black)("g"))) * "1 mole AgNO"_3/(169.87color(red)(cancel(color(black)("g")))) = "0.13245 moles AgNO"_3#
So, what would the theoretical yield of the reaction be?
Well, the
#0.13245color(red)(cancel(color(black)("moles AgNO"_3))) * "1 mole AgBr"/(1color(red)(cancel(color(black)("mole AgNO"_3)))) = "0.13245 moles AgBr"#
To calculate how many grams of silver bromide would contain this many moles? Once again, use the compound's molar mass
#0.13245color(red)(cancel(color(black)("moles"))) * "187.77 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("24.9 g AgBr")#
Now, a reaction's percent yield is defined as
#color(blue)(%"yield" = "actual yield"/"theoretical yield" xx 100)#
If the actual yield is
#"% yield" = (22.0color(red)(cancel(color(black)("g"))))/(24.9color(red)(cancel(color(black)("g")))) xx 100 = color(green)("88.4%")#
Here's a photo of how the silver bromide precipitate would look like